What is the area of the triangle whose vertices are (0, 0), (14, 6), and (12, 9)?

I realize that you could use Distance formula, followed by Hero's Formula to obtain the answer, but i'm sure there is a faster way to go about this (that way has too many radicals and fractions, and is easy to mess up on).

Suggestions to get me started?

Mar 25th 2010, 07:59 PM

Soroban

Hello, rtblue!

As a matter of fact, there is a neat formula for this problem.

Quote:

What is the area of the triangle whose vertices are (0, 0), (14, 6), and (12, 9)?

If you understand determinants, you'll like this . . .

The area of the triangle with vertices: . is:

. . . . .

For your three vertices:

. .

. . . .

Note: The "determinant bars" also represent absolute value.. . . . .The determinent can turn out to be negative.

Edit: correct a silly oversight . . . .

Mar 25th 2010, 08:13 PM

rtblue

Thanks, I understand now. (although you did forget to divide by 2 at the end (Wink) )